91Ó°ÊÓ

The bone density test scores are normally distributed.

The mean score is $\mathrm{μ}=0$.

The standard deviation is$\mathrm{σ}=1$.

The z-score is provided as -3.05.

Let x represent the bone density test score.

Asthe mean and standard deviation are0 and 1, respectively,x follows a standard normal distribution.

Steps to make a normal curve:

Step 1: Make a horizontal and a vertical axis.

Step 2: Mark the points -4, -2, 0, 2, and 4 on the horizontal axis and points 0.1, 0.2, 0.3, and 0.4 on the vertical axis.

Step 3: Provide titles to the horizontal and vertical axes as ‘z’ and ‘f(z)’, respectively.

Step 4: Shade the region right to z=-3.05.

The shaded area represents the probability.

Using table A-2, the area to the left of 3.05is obtained from the table in the intersection cell with the row value 3 and the column value 0.05, which is obtained as 0.9989.

The probability that the bone density score is greater than -3.05 is computed as follows.

$$\begin{gathered} P\left(z>-3.05\right)=P\left(z<3.05\right) \\ =0.9989 \end{gathered}$$

Thus, the probability that the bone density score is greater than -3.05 is 0.9989.